Question

A firm produces x tonnes of output at a total cost of C(x) = ${\displaystyle \frac{1}{10}{x}^3-4x^2-20x+7}$ find the

1. average cost
2. average variable cost
3. average fixed cost
4. marginal cost and
5. marginal average cost.

Answer 1

c(x) = f(x) + x

c(x) = ${\displaystyle \frac{1}{10}{x}^3-4x^2-20x+7}$

Then f(x) = ${\displaystyle \frac{1}{10}{x}^3-4x^2-20x}$ and k = 7

(i) Average Cost (AC) = ${\displaystyle \frac{\text{Total cost}}{\text{Output}}=\frac{\text{C}\left(x\right)}{x}=\frac{\text{f}\left(x\right)+k}{x}}$

${\displaystyle =\frac{\frac{1}{10}{x}^3-4x^2-20x+7}{x}}$

${\displaystyle =\frac{1}{10}\frac{x^3}{x}-\frac{4x^2}{x}-\frac{20x}{x}+\frac{7}{x}}$

${\displaystyle =\frac{1}{10}{x}^2-4x-20+\frac{7}{x}}$

(ii) Average Variable Cost (AVC) = ${\displaystyle \frac{\text{Variable cost}}{\text{Output}}=\frac{\text{f}\left(x\right)}{x}}$

${\displaystyle =\frac{\frac{1}{10}{x}^3-4x^2-20x}{x}}$

${\displaystyle =\frac{1}{10}\frac{x^3}{x}-\frac{4x^2}{x}-\frac{20x}{x}}$

${\displaystyle =\frac{1}{10}{x}^2-4x-20}$

(iii) Average Fixed Cost (AFC) = ${\displaystyle \frac{\text{Fixed cost}}{\text{Output}}=\frac{k}{x}=\frac{7}{x}}$

(iv) Marginal Cost (MC) = ${\displaystyle \frac{\text{dC}}{\text{dx}}}$

${\displaystyle =\frac{\text{d}}{\text{dx}}(\frac{1}{10}{x}^3-4x^2-20x+7)}$

${\displaystyle =\frac{\text{d}}{\text{dx}}\left(\frac{1}{10}{x}^3\right)-\frac{\text{d}}{\text{dx}}\left(4x^2\right)-\frac{\text{d}}{\text{dx}}\left(20x\right)+\frac{\text{d}}{\text{dx}}\left(7\right)}$

${\displaystyle =\frac{1}{10}\frac{\text{d}}{\text{dx}}\left(x^3\right)-4\frac{\text{d}}{\text{dx}}\left(x^2\right)-20\frac{\text{d}}{\text{dx}}\left(x\right)+0}$

= ${\displaystyle \frac{1}{10}\left(3\right)x^{3-1}-4{\left(2\right)}^{2-1}-20\left(1\right)}$

${\displaystyle =\frac{3}{10}{x}^2-8x-20}$

(v) Marginal Average Cost (MAC) = ${\displaystyle \frac{\text{d}}{\text{dx}}}$(AC)

${\displaystyle =\frac{\text{d}}{\text{dx}}(\frac{1}{10}{x}^2-4x-20)+\frac{7}{x}}$

${\displaystyle =\frac{1}{10}\frac{\text{d}}{\text{dx}}\left(x^2\right)-4\frac{\text{d}}{\text{dx}}\left(x\right)-\frac{\text{d}}{\text{dx}}\left(20\right)+\frac{\text{d}}{\text{dx}}\left(\frac{7}{x}\right)}$

${\displaystyle =\frac{1}{10}\left(2x^{2-1}\right)-4\left(1\right)-0-\frac{7}{x^2}}$

${\displaystyle =\frac{1}{5}x-4-\frac{7}{x^2}}$